A circular coil of 16 turns and radius 10 ;cm | vedclass

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Question

(D) Number of turns in the circular coil,$N = 16$.Radius of the coil,$r = 10 \;cm = 0.1 \;m$.Area of the coil,$A = \pi r^2 = \pi 	imes (0.1)^2 = 0.01

Vedclass · Accepted Answer

$1.19 	imes 10^{-4} \;kg \;m^2$

Vedclass · Answer

(D) Number of turns in the circular coil,$N = 16$.Radius of the coil,$r = 10 \;cm = 0.1 \;m$.Area of the coil,$A = \pi r^2 = \pi 	imes (0.1)^2 = 0.01\pi \;m^2$.Current in the coil,$I = 0.75 \;A$.Magnetic field strength,$B = 5.0 	imes 10^{-2} \;T$.Frequency of oscillations,$
u = 2.0 \;s^{-1}$.Magnetic moment,$M = N I A = 16 	imes 0.75 	imes 0.01\pi = 0.12\pi \;A \;m^2 \approx 0.377 \;J \;T^{-1}$.The frequency of oscillation for a magnetic dipole in a magnetic field is given by $
u = \frac{1}{2\pi} \sqrt{\frac{MB}{I_{rot}}}$,where $I_{rot}$ is the moment of inertia.Rearranging for $I_{rot}$: $I_{rot} = \frac{MB}{4\pi^2 
u^2}$.Substituting the values: $I_{rot} = \frac{0.377 	imes 5.0 	imes 10^{-2}}{4 	imes \pi^2 	imes (2.0)^2}$.$I_{rot} = \frac{0.01885}{157.91} \approx 1.19 	imes 10^{-4} \;kg \;m^2$.

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